思路：题意转化为求 (ax+by=dis) || (ax+cy=dis) || (bx+cy=dis) 三个式子有解时的最小|x| + |y|。显然求解特解x，y直接用扩展欧几里得，那么怎么求|x| + |y|？xy关系为一条直线，那么|x| + |y|应该是在x取0或者y取0的时候，但是要整数，所以只能在周围取几个点。我们知道x=x1+b/gcd*t，那么x1+b/gcd*t = 0可以解得 t = -x1 * gcd / b。然后在附近取几个点。
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define LS(n) node[(n)].ch[0]
#define RS(n) node[(n)].ch[1]
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int maxn = + ; ll ex_gcd(ll a, ll b, ll &x, ll &y){
ll d, t;
if(b == ){
x = ;
y = ;
return a;
}
d = ex_gcd(b, a%b, x, y);
t = x-a/b*y;
x = y;
y = t;
return d;
} ll dis;
//求|x|+|y|最小
ll solve(ll a, ll b){
ll x, y, d = ex_gcd(a, b, x, y);
if(dis % d != ) return INF;
x = x * dis / d;
y = y * dis / d;
a /= d, b /= d;
ll ans = abs(x) + abs(y); ll k;
k = -x / b - ;
for(int i = ; i <= ; i++){
ans = min(ans, abs(x + b * (k + i)) + abs(y - a * (k + i)));
} k = y / a - ;
for(int i = ; i <= ; i++){
ans = min(ans, abs(x + b * (k + i)) + abs(y - a * (k + i)));
} return ans;
}
int main(){
int T;
scanf("%d", &T);
while(T--){
ll a, b, A, B;
scanf("%lld%lld%lld%lld", &A, &B, &a, &b);
dis = abs(A - B);
ll ans;
ans = min(solve(a, a + b), min(solve(a, b), solve(b, a + b)));
printf("%lld\n", ans == INF? - : ans);
}
return ;
}